Download Homework ! Sec. 2.3, Page 24

Homework - Sec. 2.3, Page 24
Fall 2016
Pat Rossi
Name
1. If ajb; show that ( a) jb; aj ( b) ; and ( a) j ( b) :
Let the hypothesis be given (i.e., suppose that ajb):
Then 9 an integer m such that b = am:
) b = ( a) ( m) ) ( a) jb:
Also, the fact that b = am implies that ( b) = a ( m) ; hence, aj ( b) :
Finally, the fact that b = am implies that ( b) = ( a) (m) ; hence, ( a) j ( b) :
2. Given integers a; b; c; d; verify the following:
(a) If ajb: then ajbc:
Let the hypothesis be given (i.e., suppose that ajb):
Then 9 an integer m such that b = am:
) bc = (am) c = a (mc)
i.e., bc = a (mc) :
Hence, ajbc:
(b) If ajb and ajc; then a2 jbc:
Let the hypothesis be given (i.e., suppose that ajb and ajc):
Then 9 an integer(s) m and n such that b = am and c = an:
Then bc = (am) (an) = a2 (mn) :
i.e., bc = a2 (mn) ; and hence, a2 jbc:
(c) ajb if and only if acjbc; where c 6= 0:
ajb ) acjbc
Suppose that ajb: Then 9 an integer m such that b = am:
Observe: bc = (am) c = (ac) m
i.e., bc = (ac) m; and hence, acjbc:
acjbc ) ajb
Suppose that acjbc: Then 9 an integer m such that bc = (ac) m:
) bc = (am) c:
Since c 6= 0; we can divide both sides by zero.
) b = am:
Hence, ajb:
(d) If ajb and cjd; then acjbd
Let the hypothesis be given (i.e., suppose that ajb and cjd):
Then 9 an integer(s) m and n such that b = am and d = cn:
Then bd = (am) (cn) = (ac) (mn) :
i.e., bd = (ac) (mn) ; and hence, acjbd:
3. Prove or disprove: If aj (b + c) ; then either ajb or ajc:
This is false. To show that it’s false, we need to o¤er a counterexample.
Consider: a = 2; b = 3; and c = 5:
aj (b + c) ; but a - b and a - c:
5. Prove that for any integer a; one of the integers, a; a + 2; a + 4 is divisible by 3:
Let a be any integer. Then by the division algorithm, there are exactly three mutually
exclusive and mutually exhaustive possibilities — either a = 3k; a = 3k + 1; or a =
3k + 2:
Case 1: a = 3k
If a = 3k; then our assertion is proved.
Case 2: a = 3k + 1
If a = 3k + 1; then a + 2 = (3k + 1) + 2 = 3k + 3 = 3 (k + 1)
i.e., a + 2 = 3 (k + 1) : Hence, 3j (a + 2)
Case 2: a = 3k + 2
If a = 3k + 2; then a + 4 = (3k + 2) + 4 = 3k + 6 = 3 (k + 2)
i.e., a + 4 = 3 (k + 2) : Hence, 3j (a + 4)
Since all possibilities have been exhausted and since our proposition is true for each
possibility, we can say that for any integer a; one of the integers, a; a + 2; a + 4 is
divisible by 3:
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8. ~
(a) The sum of the squares of two odd integers cannot be a perfect square.
Two arbitrary odd integers can be represented as 2k + 1 and 2j + 1:
The sum of their squares is (2k + 1)2 + (2j + 1)2 = 4k 2 + 4k + 1 + 4j 2 + 4j + 1 =
4k 2 +4k +4j 2 +4j +2 = 4 (k 2 + k + j 2 + j)+2 = 4m+2; where m = k 2 +k +j 2 +j
The point here, is that the sum of the squares of any two odd numbers is of the
form 4m + 2:
Could this be a perfect square? Well, let’s see what form(s) perfect squares can
have.
An even number 2n; when squared, has the form: 4n2 = 4m; where m = n2 :
An odd number 2n + 1; when squared, has the form: 4n2 + 4n + 1 = 4m + 1; where
m = n2 + n:
What we have learned is that perfect squares are either of the form 4m or 4m + 1:
Perfect squares are never of the form 4m + 2; which is the form that the sum of
the squares of two odd numbers always has.
Hence, the sum of the squares of two odd integers cannot be a perfect square.
(b) The product of four consecutive integers is one less than a perfect square.
We can represent the four consecutive integers as n; n + 1; n + 2; n + 3:
Their product is n (n + 1) (n + 2) (n + 3) = n4 + 6n3 + 11n2 + 6n
2
This is one less than n4 + 6n3 + 11n2 + 6n + 1 = (n2 + 3n + 1) ; which is a perfect
square.
9. Establish that the di¤erence of two consecutive cubes is never divisible by 2.
Case 1: The smaller number is even.
Let a be even. Then 9 a natural number k such that a = 2k: This being the case,
a + 1 = 2k + 1
The di¤erence of the cubes of these numbers is (a + 1)3
12k 2 + 6k + 1 = 2 6k 2 + 3k + 1; which is odd.
{z
}
|
a3 = (2k + 1)3
(2k)3 =
2m+1
Case 2: The smaller number is odd.
Let a be odd. Then 9 a natural number k such that a = 2k + 1: This being the case,
a + 1 = 2k + 2
The di¤erence of the cubes of these numbers is (a + 1)3
12k 2 + 18k + 7 = 2 6k 2 + 9k + 3 + 1; which is odd.
|
{z
}
a3 = (2k + 2)3
(2k + 1)3 =
2m+1
Since this exhausts all cases, and each case results in the di¤erence of two consecutive
cubes being odd, our assertion is proved.
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11. If a and b are integers, not both of which are zero, verify that:
gcd (a; b) = gcd ( a; b) = gcd (a; b) = gcd ( a; b) :
d = gcd (a; b)
, d is the smallest natural number such that 9x; y 2 Z such that ax + by = d
, d is the smallest natural number such that9x1 ; y 2 Z such that
(Eq.1)
( a) x1 + by = d
| {z }
x1 =
x
(and consequently d = gcd ( a; b)
, d is the smallest natural number such that 9x1 ; y1 2 Z such that
( a) x + ( b) y = d
| {z }1 | {z }1
x1 =
x
y1 =
y
(and consequently d = gcd ( a; b)
, d is the smallest natural number such that 9x; y1 2 Z such that (a) x +( b) y1 = d:
|{z} | {z }
x=
x1
y1 =
y
(and consequently d = gcd (a; b)
Remark 1 In Eq. 1, if d were NOT the smallest natural number such that
( a) x + by = d; then there must be a smaller natural number c such that
| {z }1
x1 =x
( a) x + by = c:
| {z }1
x1 =x
But this would imply that ax1 + by = c; contradicting the assumption that d was the
|{z}
x1 =x
smallest such natural number.
Remark 2 This was not the solution that I originally had. Julian Allagan showed me
this proof and I liked it, so I changed the one that I had.
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18a. Prove: The product of any three consecutive integers is divisible by 6:
Let the integers be represented as n; n + 1; n + 2:
Claim: At least one of the integers must be even.
By the Division Algorithm, n is either of the form 2k or 2k + 1:
If n is of the form 2k; our claim is proved.
If n is of the form 2k + 1; then n + 1 = (2k + 1) + 1 = 2 (k + 1) ; and hence, n + 1 is
even.
End of Claim
Claim: At least one of the integers must be divisible by 3:
By the Division Algorithm, n is either of the form 3k; 3k + 1; or 3k + 2:
If n is of the form 3k; then our claim is proved.
If n is of the form 3k + 1; then n + 2 = (3k + 1) + 2 = 3 (k + 1) ; and hence, n + 2 is
divisible by 3:
If n is of the form 3k + 2; then n + 1 = (3k + 2) + 1 = 3 (k + 1) ; and hence, n + 1 is
divisible by 3:
End of Claim
Thus 2j (n) (n + 1) (n + 2) and 3j (n) (n + 1) (n + 2) :
Since gcd (2; 3) = 1; the second corollary to Theorem 2.4 tells us that
(2 3) j (n) (n + 1) (n + 2) :
i.e., the product of any three consecutive integers is divisible by 6:
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18b. Prove: The product of any four consecutive integers is divisible by 24:
Let the integers be represented as n; n + 1; n + 2; and n + 3:
By our results in part 11a, the product (n) (n + 1) (n + 2) (n + 3) is divisible by 3:
Claim: Our product is divisible by 8:
By the Division Algorithm, n is either of the form 4k; 4k + 1; 4k + 2; or 4k + 3:
If n is of the form 4k; then 4k+2 = 2 (2k + 1) ; and the product (n) (n + 1) (n + 2) (n + 3) =
(4k) (4k + 1) (4k + 2) (4k + 3) = (4k) (4k + 1) 2 (2k + 1) (4k + 3)
= 8k (4k + 1) (2k + 1) (4k + 3) : Our claim is proved.
If n is of the form 4k + 1; then n + 1 = (4k + 1) + 1 = 2 (2k + 1) : Furthermore,
n + 3 = (4k + 1) + 3 = 4 (k + 1) :
Hence, our product (n) (n + 1) (n + 2) (n + 3) = (4k + 1) (4k + 2) (4k + 3) (4k + 4) =
(4k + 1) 2 (2k + 1) (4k + 3) 4 (k + 1) = 8 (4k + 1) (2k + 1) (4k + 3) (k + 1) :
And our claim is proved.
Similar arguments can be used to show that if n = 4k + 2 or n = 4k + 3; then
8j (n) (n + 1) (n + 2) (n + 3)
End of Claim
We have established that 8j (n) (n + 1) (n + 2) (n + 3) and that 3j (n) (n + 1) (n + 2) (n + 3)
Since gcd (3; 8) = 1; the second corollary to Theorem 2.4 tells us that
(3 8) j (n) (n + 1) (n + 2) (n + 3) :
i.e., the product of any four consecutive integers is divisible by 24:
18c. Prove: The product of any …ve consecutive integers is divisible by 120:
By our results in part 11b, the product (n) (n + 1) (n + 2) (n + 3) (n + 4) is divisible
by 24:
Since gcd (5; 24) = 1; it remains to show that 5j (n) (n + 1) (n + 2) (n + 3) (n + 4) :
An argument similar to those of part 11a., can be used to show this.
Hence, the product of …ve consecutive integers is divisible by 120:
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